Problem: Divide the following complex numbers. $ \dfrac{-10+2i}{3+2i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3-2i}$ $ \dfrac{-10+2i}{3+2i} = \dfrac{-10+2i}{3+2i} \cdot \dfrac{{3-2i}}{{3-2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-10+2i) \cdot (3-2i)} {(3+2i) \cdot (3-2i)} = \dfrac{(-10+2i) \cdot (3-2i)} {3^2 - (2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-10+2i) \cdot (3-2i)} {(3)^2 - (2i)^2} = $ $ \dfrac{(-10+2i) \cdot (3-2i)} {9 + 4} = $ $ \dfrac{(-10+2i) \cdot (3-2i)} {13} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-10+2i}) \cdot ({3-2i})} {13} = $ $ \dfrac{{-10} \cdot {3} + {2} \cdot {3 i} + {-10} \cdot {-2 i} + {2} \cdot {-2 i^2}} {13} $ Evaluate each product of two numbers. $ \dfrac{-30 + 6i + 20i - 4 i^2} {13} $ Finally, simplify the fraction. $ \dfrac{-30 + 6i + 20i + 4} {13} = \dfrac{-26 + 26i} {13} = -2+2i $